orthogonal group of an inner product space




Given an inner product space V=(V,,)V = (V,\langle-,-\rangle), the orthogonal group of VV is the subgroup of the general linear group GL(V)GL(V) which leaves invariant the inner product.


Given an element AA of GL(V)GL(V) we say it preserves the inner product ,\langle-,-\rangle if Av,Aw=v,w\langle A v ,A w \rangle = \langle v,w \rangle for all v,wVv,w\in V.


If AA and BB preserve the inner product on VV, then so do ABAB and A 1A^{-1}.

We thus see that we have a group.


The orthogonal group O(V,,)O(V,\langle-,-\rangle) is the subgroup of GL(V)GL(V) consisting of those elements that preserve the inner product.

(… more detail …)

As a Lie group

When the base field is \mathbb{R}, \mathbb{C} or \mathbb{H} (the last being a skewfield, but this is ok), and the vector space is finite-dimensional then we can put the structure of a finite-dimensional manifold on O(V,,)O(V,\langle-,-\rangle). We shall denote the group in this case by O(n,𝕂)O(n,\mathbb{K}) where V=𝕂 nV = \mathbb{K}^n, 𝕂{,,}\mathbb{K} \in \{ \mathbb{R},\mathbb{C},\mathbb{H}\} and we take the ‘standard’ inner products on these spaces.

A relatively easy way to see that O(n,𝕂)O(n,\mathbb{K}) is a manifold is that it is a smooth affine variety in Euclidean space, but this requires some machinery. We can however construct explicit charts for O(n,𝕂)O(n,\mathbb{K}) as a real manifold.

Charts on the underlying manifold

One can use the algebra structure on matrices over the base field (or base division ring) of VV to define charts which are different to the charts one gets via the exponential map.

For AO(n,𝕂)A \in O(n,\mathbb{K}) Define the sets

Ω(A):={XO(n,𝕂)|A+XGL(V)} \Omega(A) := \{ X\in O(n,\mathbb{K}) | A + X \in GL(V) \}


T A :={YM n(𝕂)|AY+(AY) =0}. T_{A^\dagger} := \{ Y\in M_n(\mathbb{K}) | AY + (AY)^\dagger = 0 \}.

Here T A T_{A^\dagger} is a vector space, and will turn out to be the tangent space to O(n,𝕂)O(n,\mathbb{K}) at A A^\dagger.


There is an isomorphism c A:Ω(A)T A c_A\colon \Omega(A) \stackrel{\sim}{\to} T_{A^\dagger}.


The map c Ac_A is defined as

X(IA X)(A+X) 1=(IA X)(I+A X) 1A X \mapsto (I - A^\dagger X)(A + X)^{-1} = (I - A^\dagger X)(I + A^\dagger X)^{-1}A^\dagger

the latter equality using that A 1=A A^{-1} = A^\dagger. We need to check that this is indeed a map to T A T_{A^\dagger} and that it is an isomorphism.


(Ac A(X)) =Ac I(A X) A , (A c_A(X))^\dagger = A c_I(A^\dagger X)^\dagger A^\dagger,

if we show that c I(A X) =c I(A X)c_I(A^\dagger X)^\dagger = - c_I(A^\dagger X) we are done. Let Z=X AZ = X^\dagger A, which is again in O(n,𝕂)O(n,\mathbb{K}). Then

c I(Z) =((IZ)(I+Z) 1) =((IZ)Z Z 1(I+Z) 1) =((IZ )(I+Z ) 1) =(I+Z) 1(IZ) \begin{aligned} c_I(Z)^\dagger & = \left( (I - Z)(I + Z)^{-1} \right)^\dagger\\ & = \left( (I - Z)Z^\dagger Z^{\dagger -1}(I + Z)^{-1} \right)^\dagger\\ & = \left( -(I - Z^\dagger)(I + Z^\dagger)^{-1} \right)^\dagger \\ & = -(I + Z)^{-1}(I - Z) \end{aligned}



  • Euclidean space n\mathbb{R}^n with the standard inner product x,y=xy\langle\mathbf{x},\mathbf{y}\rangle = \mathbf{x}\cdot\mathbf{y} has as orthogonal group the standard orthogonal group O(n)O(n).

  • The finite-dimensional Hilbert space n\mathbb{C}^n with the standard inner product has as orthogonal group U(n)U(n), the unitary group

  • The space n\mathbb{H}^n, for \mathbb{H} the quaternions, has an inner product … such that the corresponding orthogonal group is the compact symplectic group Sp(n)Sp(n).

  • (…Examples of indefinite signature go here…)

Last revised on August 14, 2013 at 13:05:31. See the history of this page for a list of all contributions to it.